Chứng minh rằng $\frac{1-2sin^{2}2x}{1-sin4x}$ = $\frac{1+tan2x}{1-tan2x}$

$VP=\dfrac{1+\dfrac{\sin2x}{\cos2x} }{1-\dfrac{\sin2x}{\cos2x}}$

$=\dfrac{\dfrac{\cos2x+\sin2x}{\cos2x} }{\dfrac{\cos2x-\sin2x}{\cos2x}}$

$=\dfrac{\cos2x+\sin2x}{\cos2x-\sin2x}$

$=\dfrac{\cos^22x-\sin^22x}{(\cos2x-\sin2x)^2}$

$=\dfrac{\cos4x}{\cos^22x+\sin^22x-2\sin2x\cos2x}$

$=\dfrac{1-2\sin^22x}{1-\sin4x}$

$=VT$